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Calculus and Beyond Homework Help
Show that a recursive sequence converges
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[QUOTE="Mr Davis 97, post: 6056194, member: 515461"] Here is my argument: (1) WTS: ##\forall n \ge 1##, ##a_n \ge \sqrt{x}##. Note that it is true that ##(x - y^2)^2 \ge 0##. This implies that ##x^2-2xy^2+y^4 \ge 0 \implies x^2 + y^4 \ge 2y^2x \implies x+y^2 \ge 2y\sqrt{x} \implies \frac{\frac{x}{y} + y}{2} \ge \sqrt{x} \implies a_1 \ge \sqrt{x}##. So the base case is satisfied. Now, suppose that for some ##k \in \mathbb{N}## we have that ##a_k \ge \sqrt{x}##. Then ##a_{k+1} = \frac{\frac{x}{a_k} + a_k}{2} \ge \frac{\frac{x}{\sqrt{x}} + \sqrt{x}}{2} = \sqrt{x}##. This completes the induction. (2) WTS: ##\forall n \ge 1##, ##a_n \ge a_{n+1}##. We see that ##a_{n+1} = \frac{\frac{x}{a_n}+a_n}{2} \le \frac{\frac{a^2_n}{a_n}+a_n}{2} = a_n##. So we see that ##(a_n)## is bounded below and is decreasing, so by the MCT it has a limit, L. Then ##L = \frac{\frac{x}{L}+L}{2} \implies L = \sqrt{x}##. [/QUOTE]
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Show that a recursive sequence converges
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